golang编程题

1、开启100个协程，顺序打印1-1000，并保证协程号为1的，打印尾数也是1

package main

import "fmt"

const (
	routineNum = 100
	numMax     = 1000
)

func main() {
	s := make(chan struct{})
	channels := make(map[int]chan int, 100)
	for i := 1; i <= routineNum; i++ {
		channels[i] = make(chan int)
	}
	for i := 1; i <= routineNum; i++ {
		go func(idx int) {
			for {
				num := <-channels[idx]
				fmt.Printf("chan id:[%d],num:[%d]\n", idx, num)
				s <- struct{}{}
			}

		}(i)
	}

	for j := 1; j <= numMax; j++ {
		idx := j % routineNum //0~99
		if idx == 0 {
			idx = routineNum
		}
		channels[idx] <- j
		<-s
	}
	return
}

2、3个goroutinue交替打印abc10次

package main

import (
	"fmt"
	"sync"
)

func main() {
	chan1 := make(chan struct{})
	chan2 := make(chan struct{})
	chan3 := make(chan struct{})
	var wg sync.WaitGroup
	wg.Add(3)

	go func() {
		defer wg.Done()
		for i := 0; i < 10; i++ {
			<-chan1
			fmt.Println("a")
			chan2 <- struct{}{}
		}
		<-chan1
	}()
	go func() {
		defer wg.Done()

		for i := 0; i < 10; i++ {
			<-chan2
			fmt.Println("b")
			chan3 <- struct{}{}
		}

	}()
	go func() {
		defer wg.Done()

		for i := 0; i < 10; i++ {
			<-chan3
			fmt.Println("c")
			chan1 <- struct{}{}
		}
	}()
	chan1 <- struct{}{}
	wg.Wait()

	close(chan1)
	close(chan2)
	close(chan3)
}

3、用不超过10个goroutinue打印不重复的打印slice中100个元素

4、用2个协程交替打印奇偶数